Page 388 - Schaum's Outline of Theory and Problems of Advanced Calculus
P. 388

CHAP. 15]                  GAMMA AND BETA FUNCTIONS                             379


                                                           ð 1
                                                              x 1    y 1
                                                             t          dt
                                                   Bðx; yÞ¼     ð1   tÞ                             ð16Þ
                                                            0
                        If x   1 and y   1, this is a proper integral. If x > 0; y > 0 and either or both x < 1or y < 1, the
                     integral is improper but convergent.
                        It is shown in Problem 15.11 that the beta function can be expressed through gamma functions in the
                     following way
                                                               ðxÞ  ð yÞ
                                                       Bðx; yÞ¼                                     ð17Þ
                                                                ðx þ yÞ
                        Many integrals can be expressed through beta and gamma functions.  Two of special interest are
                                          ð  =2                 1
                                              sin 2x 1    cos 2y 1    d  ¼ Bðx; yÞ¼  1  ðxÞ  ð yÞ   ð18Þ
                                           0                    2        2  ðx þ yÞ
                                         ð    p 1
                                          1  x
                                                                           0 < p < 1                ð19Þ
                                                 dx ¼  ð pÞ  ð p   1Þ¼
                                          0 1 þ x                  sin  p
                     See Problem 15.17.  Also see Page 377 where a classical reference is given.  Finally, see Chapter 16,
                     Problem 16.38 where an elegant complex variable resolution of the integral is presented.

                     DIRICHLET INTEGRALS
                                                                                    x p  y q   z r

                        If V denotes the closed region in the first octant bounded by the surface  þ  þ  ¼ 1 and
                                                                                    a    b     c
                     the coordinate planes, then if all the constants are positive,



                                        ðð ð
                                               1    1 
 1       a b c   p    q    r
                                            x   y   z  dx dy dz ¼                                   ð20Þ
                                                                 pqr
                                         V                              1 þ  þ  þ
                                                                           p   q  r
                        Integrals of this type are called Dirichlet integrals and are often useful in evaluating multiple
                     integrals (see Problem 15.21).


                                                     Solved Problems

                     THE GAMMA FUNCTION

                     15.1. Prove:  (a)  ðx þ 1Þ¼ x ðxÞ; x > 0;  ðbÞ  ðn þ 1Þ¼ n!; n ¼ 1; 2; 3; .. . .
                                                     M
                                      ð             ð
                                      1  v  x           v  x
                                        x e  dx ¼ lim  x e  dx
                           ðaÞ  ðv þ 1Þ¼
                                      0          M!1  0
                                                      ð M
                                                           x
                                            v
                                                 x M
                                    ¼ lim  ðx Þð e Þj 0    ð e Þðvx v 1 Þ dx
                                      M!1             0
                                                     M
                                                    ð
                                             v  M
                                    ¼ lim   M e   þ v  x v 1  x  dx ¼ v ðvÞ  if v > 0
                                                          e
                                      M!1            0
                                   ð            ð M
                                    1
                                     e  x  dx ¼ lim  e  x  dx ¼ lim ð1   e  M Þ¼ 1:
                           ðbÞ  ð1Þ¼
                                    0       M!1  0       M!1
                                  Put n ¼ 1; 2; 3; .. . in  ðn þ 1Þ¼ n ðnÞ.  Then
                                         ð2Þ¼ 1 ð1Þ¼ 1;  ð3Þ¼ 2 ð2Þ¼ 2   1 ¼ 2!;  ð4Þ¼ 3 ð3Þ¼ 3   2! ¼ 3!
                                  In general,  ðn þ 1Þ¼ n! if n is a positive integer.
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